SOLUTIONS TO ASSIGNMENT 1 MATH 576

SOLUTIONS TO ASSIGNMENT 1 MATH 576 SOLUTIONS BY OLIVIER MARTIN 13 #5. Let T be the topology generated by A on X. We want to show T = J B J where B is the set of all topologies J on X with A J. This amounts to showing that if A J then T J. This is automatic since J being a topology arbitrary unions of elements of J, and thus arbitrary unions of elements of A, lie in J. But Lemma 13.1 gives us that T consists of all unions of elements of A so that T J. In short, the topology generated by the basis A is minimal in the sense that any topology containing A will contain it. For the second part, namely for subbases, one uses the same argument. If A J then since J is topology, finite intersections of elements of A lie in J and thus arbitrary unions of finite intersections of elements of A also lie in J. Since T, the topology generated by the subbasis A consists precisely of these elements by the definition of the topology generated by a subbasis T J and hence T satisfies the same kind of minimality condition as T. #7. T 1 contains T 3 and T 5. Indeed consider A = R {a 1, a 2,..., a n } T 3 then A T 1 since A = (, a 1 ) (a 1, a 2 )... (a n, ). Moreover the basis generating T 5 is a subset of T 1 hence so is T 5. T 2 contains T 1 since the basis generating T 2 contains the basis generating T 1, T 3 since T 3 T 1 and T 5 since T 5 T 1. T 3 contains no other topology. T 4 contains T 1 and thus T 3 and T 5 since given a basis elements (a, b) T 1 and x (a, b) then (a, x] (a, b). Also, T 2 is a subset of T 4. Indeed, all basis elements of the form (a, b) are in T 4 since T 1 is. Given a basis element B of the form (a, b) K and x ((a, b) K) (0, ), let c be the greatest rational of the form 1/n which is less than x and let b = (c + x)/2. Then (b, x] is a basis element of T 4 containing x and which is contained in B. If x 0, (a, x] is also a basis element containing x and contained in B. Hence T 4 is finer than T 2. Finally, T 5 does not contain any of the other topologies. 1

#8. a) Consider (R, T ) where T is the standard topology on the real line. To establish that B is a basis we need to show that for any open set U in T and any x U, there exists a B B with x B U. This is obvious since, the set of all open interval being a basis for T, U = i I A i where the A i are open intervals and I is some index set. Then if x U, x A i for some A i, say (a, b). Then letting B = (c, d) where c and d are rationals lying respectively in (a, x) and (x, b) and which exists by the fact that Q is dense in R, we have as required x B U with B B. b) We only need to exhibit an open subset of R l which cannot be expressed as a union of elements of C, thereby showing that the topology generated by C is not equal to the lower limit topology on the real line. We claim that [π, 4) is such an open subset. Indeed, assume [π, 4) = i I C i with C i C. Then π C i for some i I and thus C i = [π, b), a contradiction since π Q. 16 #1. Let T be the topology on X then the subspace topology on A as a subspace of X is given by intersections of elements of T with A while the topology it inherit from the topology on Y consists of the intersection of the elements of T intersected with Y with A. But given B T, (B Y ) A = B A since A Y. #2. Obviously C, the subspace topology induced by T, is finer than C, the subspace topology induced by T, since all intersections of elements of C and Y are intersections of elements of C and Y. Yet C need not be strictly finer than C. Indeed, take X = {a, b, c, d}, T = {, {a}, {b}, {a, b}, X}, T = {, {a, b}, X} and Y = {c, d}. #10. Let T 1 be the product topology on I I, T 2 the dictionary order topology and T 3 the subspace topology I I inherits as a subspace of R R with the dictionary order topology. T 1 is not comparable with T 2. Indeed, consider the basis element A = [0, 1/2) (1/2, 1] of T 1. The point x with coordinates (1/4, 1) belongs to A yet no basis element of T 2 containing x is a subset of A since basis elements containing x contain elements greater than x but whose second coordinate is less than 1/2. Conversely, the basis element B = [1/2, 1/2] (1/4, 1/2) clearly cannot contain any basis element of T 1 (since these are intersections of elements of the form (a, b) (c, d) with I I because the product topology on I I is equal to the subspace topology). Hence choosing any x in B and using lemma 13.3 shows these two topologies are not comparable. Moreover T 1 is strictly coarser than T 3. Indeed,

given a basis element C of T 1 and some x in C we can show there is an element of a basis for T 3 containing x which is a subset of C. We have a two cases to check. If x = (x 1, x 2 ) [0.1] (0, 1) then there exists some ɛ such that A := [x 1, x 1 ] (x 2 ɛ, x 2 + ɛ) C with A T 3 otherwise C would be the intersection of a rectangle which is the product of some interval with an interval (c, x 2 ], [c, x 2 ], [x 2, c) or [x 2, c] and such a subset is not open in the subspace topology T 1. If x [0, 1] (0, 1) then its second coordinate is 0 or 1 and in the first we can choose some ɛ > 0 such that D := [x 1, x 1 ] [x 2, x 2 + ɛ) C with D T 3 whereas in the second case one can similarly choose ɛ > 0 and D := [x 1, x 1 ] (x 2 ɛ, x 2 ] C. Consequently, T 1 T 3. This inclusion is strict since for the open subset B = [1/2, 1/2] (1/4, 1/2) and the point x = (1/2, 3/8) in this subset, there is clearly no basis element for T 1 containing x and which is a subset of B. Finally, we need to compare T 3 and T 2. We claim that T 3 in strictly finer than T 2. Indeed the open subset D = [1/2, 1/2] (1/2, 1] in T 3 is not open in T 2 since any subset containing (1/2, 1) will also contain points whose first coordinate greater than 1/2. Finally T 2 T 3 since any basis element for T 2 is also a basis element for T 3. 17 # 2 If A is closed in Y then by theorem 17.2 there exists U closed in X such that A = U Y. Y being closed in X, U Y and thus A is also closed in X. # 3 We want to show that X Y A B is open in the product topology. But X Y A B = (X (Y B)) ((X A) Y ), a union of two open subsets, and is thus open. To show this last equality observe that X Y A B = {(x, y) X Y : x X or y Y } = {(x, y) X Y : x X} {(x, y) X Y : y Y } = (X (Y B)) ((X A) Y ). # 5 Let A = (a, b). We need to show that any x R [a, b] is not a limit point of (a, b). If x < a, (, a) or (a 0, a), where a 0 is the smallest element of the set X if it exists, is an open set containing x in the order topology on X and it does not intersect A. Similarly, if x > b, (b, ) or (b, b 0 ), where b 0 in the greatest element of X if it exists, is an open set containing x and disjoint from A. Thus (a, b) [a, b]. Equality holds if and only if for any c > a, d < b (a, b) (a, c), (a, b) (d, b). This is equivalent to the fact that for any c > a, d < b there exist c, d such that c (a, c), d (d, b). This is equivalent to the fact that the sets {c X : c > a} and {d X : d < b} do not have

respectively minimal and maximal elements. It is thus a necessary, but not sufficient, condition that these two sets be infinite. Note that the condition derived above can be restated as the fact that a and b are limit points of (a, b). # 6 a) Use theorem 17.5. If every neighborhood of x A intersects A they clearly intersects B hence A B. b) Use theorem 17.5 again. Clearly A B A B since given x A B, ever neighborhood of x intersects A or every neighborhood of x intersects B, so in particular every neighborhood of x intersects A B, i.e. x A B. The reverse inclusion follows from the fact that A B contains A B and is closed. Hence, A B being the intersection of all closed sets containing A B, is contained in A B. c) For any x α A α, every neighborhood of x intersect with a given A α (that is some A α which works for all neighborhoods of x) hence, in particular every neighborhood of x intersects with one of the A α and thus, by theorem 17.5, x α A α. Consider R with the standard topology and consider A n = {1/n}, Then since A n is closed A n = A n and n N A n = {1/n : n N} yet n N A n = {1/n : n N} {0}. Indeed any neighborhood of 0 will contain some ɛ > 0 and there is some m N with 1/m < ɛ. This gives an example where equality fails. #9 A B is closed in the product topology and thus contains A B. For the reverse inclusion, let x be in A B. Then every neighborhood of the projection of x on X intersects A and every neighborhood of the projection of x on Y intersects B. Since every neighborhood of x in the product topology contains a basis element containing x which is the product of a neighborhood of the projection of x on X and a neighborhood of the projection on Y, it follows that every neighborhood of x intersects A B. Indeed, if a neighborhood U of the projection of x on X and a neighborhood V of the projection of x on Y, respectively intersect A and B at a and b, then U V will intersect A B at (a, b). Hence A B A B and the equality has been proved. # 11 Let X and Y be two Hausdorff spaces. Given x = (x 1, y 1 ) y = (x 2, y 2 ) X Y either x 1 x 2 or y 1 y 2. In the first case there exists A, B X open disjoint neighborhoods of x 1 and x 2 respectively. Then A Y and B Y are neighborhoods of x and y respectively and (A Y ) (B Y ) = (A B) Y =. If y 1 y 2 there exists C, D Y open disjoint neighborhoods of y 1 and y 2 respectively. Then X C and X D are neighborhoods of x and y respectively and

(X C) (X D) = X (C D) =. Therefore if X and Y are both Hausdorff so is X Y. # 12 Let Y be a subspace of a Hausdorff space X. Given x y Y there exists two disjoint neighborhoods of x and y, say U and V in X. Then U Y and V Y are open in Y and are thus also neighborhoods of x and y in Y. Moreover (U Y ) (V Y ) = (U V ) Y =. Thus Y is Hausdorff. # 13 Assume X is Hausdorff. Given x y in X there exist disjoint neighborhoods U, V of x and y. U V {(x, y) X X : x y} := A and U V is a basis element for the product topology on X X. Hence for any point in A there is a neighborhood of this point which is in A and so A is open in X X and its complement is closed. Conversely assume is closed then A is open and so for every (x, y) A there is a basis element containing (x, y) and which is a subset of A. Let this element be U V. Then since it is a subset of A, U V = and U and V are respectively neighborhoods of x and y. Since this holds for all x y in X, X is a Hausdorff topological space. # 19 a) Let x be in Int A then there is a neighborhood of x which is a subset of A, say U. Then x is not in (X A) since if it were U would intersect X A by Theorem 17.5. Now let us show A Int A Bd A. If x A then either x Int A (in which case there exists a neighborhood of x which is a subset of A) or every neighborhood of x intersects (X A) and A. This is equivalent to x A (X A). the reverse inclusion is trivial as Int A A A and Bd A A. b) If A is both open and closed A = A, (X A) = X A so that A (X A) =. If Bd A = then A = Int A Bd A = Int A so that A is open because Int(A) A A implies A = Int(A). Similarly if Bd A is empty so is Bd (X A) = Bd A so that (X A) = Int (X A) Bd (X A) = Int (X A) so that X A is open and A is closed. c) If U is open Bd U = U X U = U (X U) = U U. If Bd U = U U then Int U Bd U = (U U) Int U = U. Since Int U U we conclude Int U = U and U is open. d) If U is open U need not be equal to Int U. A simple counterexample comes from a case where U is a set which is both open and closed while U is not. Then Int U = U U. Take for instance R equipped with the finite complement topology and let U = R {0}. Then Int U = R since the closure of U is R itself which is closed and open.